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Pressure inside two soap bubbles are `1.01` and `1.02` atmospheres. Ratio between their volumes isA. 16B. 8C. 4D. 2 |
Answer» Correct Answer - B Excess pressure in one bubble, `p_(1)=1.01 - 1=0.01 =(4S)/(r_(1))` …(i) Excess pressure in other bubble, `p_(2)=1.02 - 1=0.02=(4S)/(r_(2))` ….(ii) Dividing (i) by (ii) `(0.01)/(0.02)=(r_(2))/(r_(1))` or `(r_(1))/(r_(2))=2` `(V_(1))/(V_(2))=(r_(1)^(3))/(r_(2)^(3))=(2)^(3) =8` |
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